Differential forms and vector calculus

Sean Fitzpatrick
November 11, 2008

This is a brief introduction to the language of differential forms in R3 and how they relate to
vector calculus. You may ask, why bother to re-write everything when we already have perfectly
good formulas in terms of vectors? For one, I hope you’ll agree that these formulas look a lot nicer,
and are more easily remembered, in terms of forms. Another reason is the need to do vector calculus
in spaces other than R3 . For example, electrodynamics takes place in 4D (3D space + time), and
general relativity requires curved space. Our vector calculus formulas don’t work in these settings
(try defining the cross product of two four-dimensional vectors!), but differential forms work just
fine. So let’s get started with the translation!

1

Differential forms on R3

Let f (x, y, z) be a smooth function, and let F~ (x, y, z) = P (x, y, z)ˆı + Q(x, y, z)ˆ
 + R(x, y, z)kˆ be a
smooth vector field. (For the purposes of this note, “smooth” will mean that any partial derivatives
we need exist and are continuous.) We can express the four types of differential form in R3 in terms
of f and F~ as follows.
• 0-forms A 0-form is just a smooth function, such as f (x, y, z).
• 1-forms To each vector field F~ we associate a 1-form
WF~ = P (x, y, z) dx + Q(x, y, z) dy + R(x, y, z) dz,
called the work form associated to F~ . Every 1-form is an expression of this type.
• 2-forms We can also associate F~ to a 2-form ΦF~ , called the flux form associated to F~ , and
given by
ΦF~ = P (x, y, z) dy ∧ dz + Q(x, y, z) dz ∧ dx + R(x, y, z) dx ∧ dy.
• 3-forms Finally, our function f (x, y, z) can be associated to a 3-form, given by
ρf = f (x, y, z) dx ∧ dy ∧ dz,
which we’ll call the “volume form” associated to f . (Technically, to be considered a volume
form we should demand that f (x, y, z) > 0 everywhere, but we’ll abuse the language.)

1

We remark that every 1-form is the work
form of some vector field. For any smooth functions f and g.
In particular. this tells us that dx ∧ dy = −dy ∧ dx.
Recall that the differential of a continuously differentiable function f (x. If we write out the differentials in full. we can
use the properties of the wedge product to simplify this expression. then
α ∧ β = (−1)kl β ∧ α. y. y. z)dx + fy (x.
since dy ∧ dx = −dx ∧ dy and dx ∧ dx = 0.
That is. z) = fx (x. y. for example. z)β + g(x. y. the differential of f is the work form of its gradient. y. z)γ) = f (x. We also get the identity
df = W∇f . y. and the differential d. what can we do with it? We first need to define two
operations on differential forms: the wedge product appearing in ΦF~ and ρf .For each k. y. so now that we have this notation. and every 3-form is
the volume form of some function. 5-forms.
2
.
Next. F~ is a conservative vector field if its work form is the differential of some function.we get the identity
dWF~ = Φ∇×F~ . every 2-form is the flux form of some vector field. and that dx ∧ dx = dy ∧ dy = dz ∧ dz = 0. z)α ∧ β + g(x. That
is. This
also tells us why there are no 4-forms. z) is given by
df (x. y.
2. z)dz.
dWF~ =
∂y
∂z
∂z
∂x
∂x
∂y
Now. and put
everything together. If α is a k-form. we define the differential of a 1-form α = P dx + Q dy + R dz by
dWF~ = dP ∧ dx + dQ ∧ dy + dR ∧ dz. z)α ∧ γ. If we do the same for dQ ∧ dy and dR ∧ dz. notice that the functions appearing here are the components of ∇ × F~ . etc: we have. β and γ. z)dy + fz (x. and differential forms α. y. we get
∂P
∂R
∂Q ∂P
∂R ∂Q
−
dy ∧ dz +
−
dz ∧ dx +
−
dx ∧ dy. dQ. Notice that this gives an alternative
answer to the question “When is F~ a gradient?” . and dR are the usual differentials.
The wedge product ∧ is a multiplication that takes a k-form α and an l-form β and gives a
k + l-form α ∧ β such that:
1. note that the result is a 1-form. and β is an l-form. we have
dP ∧ dx = (Px dx + Py dy + Pz dz) ∧ dx
= Pz dz ∧ dx − Py dx ∧ dy. For example.
This defines the differential d on 0-forms. we have
α ∧ (f (x. we denote the space of k-forms by Ωk (R3 ).we have F~ = ∇f if and only if WF~ = df .
OK. dy ∧ (dx ∧ dy ∧ dz) =
−(dy ∧ dy) ∧ (dx ∧ dz) = 0.
where dP .

a 1-form is integrated along a curve (defined by one parameter). since there are no 4-forms on R3 . only
the term Qy dy is going to survive. courtesy of the wedge product. and a 3-form is integrated over
3
.
Summing up. Stokes’ theorem. we get a sequence of maps
d
d
d
Ω0 (R3 ) −
→ Ω1 (R3 ) −
→ Ω2 (R3 ) −
→ Ω3 (R3 )
such that df = W∇f .
In other words. applying the differential d to a
3-form gives zero.
or ∇ · (∇ × F~ ) = 0.
a 2-form is integrated over a surface (defined by two parameters). dWF~ = Φ∇×F~ . say dQ ∧ dz ∧ dx. This gives us the two separate
identities we know from vector calculus:
d2 f = d(df ) = d(W∇f ) = Φ∇×∇f = 0. applying the differential d to a 1-form has the same effect as taking the curl of a
vector field.
Now.In other words. We have:
dΦF~ = dP ∧ dy ∧ dz + dQ ∧ dz ∧ dx + dR ∧ dx ∧ dy. almost for free!).
which tells us that ∇ × ∇f = 0.
Finally.
The basic principle for integrating a differential form is this: a k-form is something you integrate
over a k-dimensional space. The first is that differential forms behave nicely under
change of variables (we get the change of variables formula. and then work out the wedge products. etc. Putting everything together.
In this case if we look at one of the terms. and we end up with Qy dy ∧ dz ∧ dx = Qy dx ∧ dy ∧ dz. Of course. we always get zero.
divergence theorem) we covered are special cases of one general result (also called Stokes’ theorem). and
d2 WF~ = d(dWF~ ) = d(Φ∇×F~ ) = ρ∇·(∇×F~ ) = 0. we get
dΦF~ = (Px + Qy + Rz ) dx ∧ dy ∧ dz = ∇ · F~ dx ∧ dy ∧ dz = ρ∇·F~ .
2
Integration of differential forms
We now come to the use of the language of forms in the integration theorems of vector calculus. if we apply d twice in a row. The pattern is the same: we take the
differentials of the component functions. let’s apply the differential d to a 2-form ΦF~ . Jacobians. The
second is that all the different theorems (FTC for line integrals. (Here
there’s no sign change because we make an even number of swaps: dy ∧ dz ∧ dx = −dy ∧ dx ∧ dz =
−(−dx ∧ dy ∧ dz).
The third is that differential forms keep track of orientation: we get a minus sign if we switch from
right-handed coordinates to left-handed. since dz and dx are already there.
There are three advantages to using forms. Green’s theorem. the differential d has the following useful property:
d2 = d ◦ d = 0. That is. and dΦF~ = ρ∇·F~ .

and dWF~ = Φ∇×F~ . if α = ΦF~ is a 2-form.
S
∂S
The boundary of S is a closed curve C. Let’s start with a 0-form
α = f (x.
double.
Finally. y. a lengthy (but straightforward) calculation lets us show that
f (x.
C
S
which is the classical Stokes’ theorem. then
Z
Z
dα =
σ
α.
5
. w))JT (u. depending on the context. v. if (x. or triple integrals. Then we get:
Z
Z
df =
f = f (B) − f (A). We write this as ∂C = B − A to indicate that the left endpoint is assigned the
opposite orientation (this is a part of the discussion I’ve skipped). so in vector form.
Now. we get
ZZ
ZZZ
dΦF~ =
ΦF~ . this equation becomes
ZZ
(∇ × F~ ) · ~n dS =
I
(F~ · T~ ) ds. this equation reads
ZZZ
ZZ
∇ · F~ dV =
F~ · ~n dS. y. let’s denote
these by A and B. v. and σ is a region defined by
(k + 1)-parameters. z).
∂σ
where ∂σ denotes the boundary of σ.
T
∂T
The boundary of T is a closed surface S.
T
S
which is the divergence theorem.
Finally. and T is a region in space. v. z) = T (u. If α is a k-form. (We’ve written a single integral sign to stand for single. and ∂σ is given by the two end-points of the curve. z) dx ∧ dy ∧ dz = f (T (u. and S is a surface.Finally.
which lets us recover the change of variables formula. we get
Z
ZZ
dWF~ = WF~ . case-by-case. so in vector form. and dΦF~ = ρ∇·F~ . we state the general Stokes’ theorem. w) is a smooth transformation.
C
B−A
which is the Fundamental Theorem for line integrals.)
Let’s translate this theorem back into vector calculus. w) du ∧ dv ∧ dw. Then σ is a curve C. y. if α = WF~ is a 1-form.

Now what about a 2-forms? Well this should be something that eats 2 vectors and returns a
number.
In particular. then dx(~v ) = a. z components. dy(~v ) = b. ~v ) = α(~u)β(~v ) − α(~v )β(~u). u2 . The 1-forms
dx. ci. v3 i. If ~u = hu1 . let’s look at the basic 2-forms such as dx ∧ dy. If α and β are 1-forms. of a vector ~v ∈ R3 : if ~v = ha.
respectively. ~v ) = dx(~u)dy(~v ) − dx(~v )dy(~u)
. dy and dz can be thought of as the linear functions on R3 that give the x. u3 i and ~v = hv1 . b. and dz(~v ) = c.Let us end with a mention of the geometric interpretation of differential forms. we define
α ∧ β(~u.
we find
dx ∧ dy(~u. y. v2 .

.

.

u1 v1 .

.

..

= u1 v2 − v1 u2 = .

u2 v2 .

dz ∧ dx and dy ∧ dz give the areas determined by
projecting onto the xz or yz planes.
which is what we get if we project ~u and ~v onto the xy-plane. dx∧dy determines the “xy-component”
of the area spanned by two vectors.
6
. respectively. Similarly. and then compute the area of the
parallelogram spanned by these projections. In other words.